Simplify; express your answer in exponential form. Assume $k\neq 0, p\neq 0$. $\dfrac{{(k)^{3}}}{{k^{-1}p^{-5}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${k}$ to the exponent ${3}$ . Now ${1 \times 3 = 3}$ , so ${(k)^{3} = k^{3}}$ In the denominator, we can use the distributive property of exponents. ${k^{-1}p^{-5} = k^{-1}p^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(k)^{3}}}{{k^{-1}p^{-5}}} = \dfrac{{k^{3}}}{{k^{-1}p^{-5}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{3}}}{{k^{-1}p^{-5}}} = \dfrac{{k^{3}}}{{k^{-1}}} \cdot \dfrac{{1}}{{p^{-5}}} = k^{{3} - {(-1)}} \cdot p^{- {(-5)}} = k^{4}p^{5}$.